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← JEE Main 2014 · 6 Apr · Shift 1
MathematicsSingle correct+4 / -1Question 65 of 90

JEE Main 2014 · 6 Apr · Shift 1, Mathematics question 65

If α,β≠0\alpha,\beta\ne0, and f(n)=αn+βnf(n)=\alpha^n+\beta^n and
∣31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)∣=K(1−α)2(1−β)2(α−β)2,\begin{vmatrix}3&1+f(1)&1+f(2)\\1+f(1)&1+f(2)&1+f(3)\\1+f(2)&1+f(3)&1+f(4)\end{vmatrix}=K(1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2,
then K is equal to :

Choose one option
Answer and worked solution

Official answer: A

The matrix is VVTVV^T with V=[1111αβ1α2β2]V=\begin{bmatrix}1&1&1\\1&\alpha&\beta\\1&\alpha^2&\beta^2\end{bmatrix}. Entry (i,j)(i,j) is 1+αi+j−2+βi+j−21+\alpha^{i+j-2}+\beta^{i+j-2}. So
det⁡=(det⁡V)2=[(1−α)(1−β)(α−β)]2,\det=(\det V)^2=\left[(1-\alpha)(1-\beta)(\alpha-\beta)\right]^2,
and K=1K=1.

Topics:Determinants and properties
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